# Proving identities

• Mar 12th 2007, 10:15 AM
Kim2425
Proving identities
I have absolutely no idea how this works, so if you can go through these two for me maybe I can do the other 8 :(

A) x(2) + 2x -2 = (x+1)(x-1) +2(x+1)-3

b) 6x + 9 = 7(x+1) (x-2)

thank you for any help
• Mar 12th 2007, 10:52 AM
topsquark
Quote:

Originally Posted by Kim2425
I have absolutely no idea how this works, so if you can go through these two for me maybe I can do the other 8 :(

A) x(2) + 2x -2 = (x+1)(x-1) +2(x+1)-3

b) 6x + 9 = 7(x+1) (x-2)

thank you for any help

The first IS an identity, the second isn't. Are you supposed to test if the equation is an identity?

a) There is nothing to do on the LHS, so go to the RHS and simplify:
(x + 1)(x - 1) + 2(x + 1) - 3

x^2 - 1 + 2x + 2 - 3

x^2 + 2x + (-1 + 2 - 3)

x^2 + 2x - 2

Since the LHS and the RHS are identical, this equation is an identity.

b) Again, nothing to do on the LHS, so go to the RHS:
7(x + 1)(x - 2)

7(x^2 - x - 2)

7x^2 - 7x - 14

The LHS and RHS are not the same, so this is not an identity, but a "conditional equation."

-Dan
• Mar 12th 2007, 10:52 AM
spiritualfields
Quote:

6x + 9 = 7(x+1) (x-2)
Expanding the right side, and rewriting yields:

6x + 9 = 7x^2 - 7x -14

How can this possibly be an identity? Perhaps you were given a list of equations and you are supposed to pick out the ones that are identities from the ones that are not?
• Mar 12th 2007, 11:43 AM
Soroban
Hello, Kim!

You're supposed to show that the two sides are equal.
. . How hard can that be?
And you have a typo in the second one . . .

Quote:

A) .x² + 2x - 2 .= .(x + 1)(x - 1) +2(x + 1) - 3
Simplify the right side: .x² - x + x - 1 + 2x + 2 - 3 .= .x² +2x - 2

. . and it equals the let side . . . get it?

Quote:

B) . 6x + 9 .= .7(x + 1) - (x - 2)
Now you try this one . . .